Problem B. Harvest of Apples（杭电2018年多校+组合数+逆元+莫队）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6333

题目：

题意：求C(n,0)+C(n,1)+……+C(n,m)的值。

思路：由于t和n数值范围太大，所以此题查询复杂度不能太高，由组合数的将前k项求和可以推出，从而可以转换成莫队的区间查询，将n当成l，m当成r即可。此题需要注意，对于求组合数得用o(1)的方法求，也就是阶乘相除的方法，对于分母我们得求逆元，因而借助欧拉定理。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int t, block;
ll sum;
ll a[maxn], b[maxn];

struct node {
    int l, r, id;
    ll ans;
    bool operator < (const node & x) const {
        return (l - 1) / block == (x.l - 1) / block ? r < x.r : l < x.l;
    }
}ask[maxn];

ll Mod_Pow(ll x, ll n) {
    ll res = 1;
    while(n > 0) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void init() {
    a[1] = 1;
    for(int i = 2; i < maxn; i++) a[i] = a[i-1] * i % mod;
    for(int i = 1; i < maxn; i++) b[i] = Mod_Pow(a[i], mod - 2);
}

ll C(int n, int m) {
    if(n < 0 || m < 0 || m > n) return 0;
    if(m == 0 || m == n) return 1;
    return a[n] * b[n-m] % mod * b[m] % mod;
}

int main() {
    //FIN;
    init();
    scanf("%d", &t);
    block = sqrt(maxn);
    sum = 1;
    for(int i = 1; i <= t; i++) {
        scanf("%d%d", &ask[i].l, &ask[i].r);
        ask[i].id = i;
    }
    sort(ask + 1, ask + t + 1);
    for(int i = 1, l = 1, r = 0; i <= t; i++) {
        while(l < ask[i].l) sum = (2 * sum - C(l++, r) + mod) % mod;
        while(l > ask[i].l) sum = ((sum + C(--l, r)) * b[2]) % mod;
        while(r < ask[i].r) sum = (sum + C(l, ++r)) % mod;
        while(r > ask[i].r) sum = (sum - C(l, r--) + mod) % mod;
        ask[ask[i].id].ans = sum;
    }
    for(int i = 1; i <= t; i++) {
        printf("%lld
", ask[i].ans);
    }
    return 0;
}